Question

If the graph f(x)= 9x^2+37x+41/3x+5 has an oblique asymptote at y=3x+k what is the value of k

Answer 1

`k=(22)/(3)`

Explanation:

The oblique asymptote of

`f(x)=(9x^2+37x+41)/(3x+5)`,

We perform the long division as shown in the attachment.

The quotient is;

`3x+(22)/(3)`

Comparing to 3x+k

Hence the value of k is
`(22)/(3)`

Answer 2

`(22)/(3)`

Explanation:

To find out oblique asymptote we divide the polynomials using long division

To find quotient divide the first term. then multiply the answer with 3x+5 and write it down. Subtract it from the top. Repeat the process till we get remainder.

`3x+(22)/(3)`

------------------------------

`3x+5`
`9x^2+37x+41`

`9x^2+15x`

-------------------------------------(Subtract)

`22x+41`

`22x+(110)/(3)`

------------------------------------(subtract)

`(13)/(3)`

Quotient is
`3x+(22)/(3)` that is our oblique asympotote

the value of k is 22/3